If 9th term of A.P. is zero, then prove that 29th term is twice the 19

1.	If 9th term of A.P. is zero, then prove that 29th term is twice the 19th term.
Answer» Let a be the first term and d be the common difference then 9^th term = T₉ = 0 Then a + (9 – 1)d= 0 ⇒ a + 8d = 0 ….. (i) 29^th term = T₂₉ = a + 28d …(ii) 19^th term = T₁₉ = a + 18d …(iii) Putting the value of equation (i) into equation (ii) and (iii) we get T₂₉ = (a + 8 d) + 20 d ⇒ T₂₉ = 0 + 20 d ⇒ T₂₉ = 20d … (iv) ⇒ T₂₉= (a + 8 d) + 10 d ⇒ T₂₉ = 0 + 10d ⇒ T₂₉ = 10d … (v) From equation (iv) and (v), we have T₂₉ = 20d = 2 × 10d =2 × T₂₉ T₂₉ = 2T₂₉ Hence Proved.

If 9th term of A.P. is zero, then prove that 29th term is twice the 19th term.

Answer»

Let a be the first term and d be the common difference then 9^th

term = T₉ = 0 Then a + (9 – 1)d= 0

⇒ a + 8d = 0 ….. (i)

29^th term = T₂₉ = a + 28d …(ii)

19^th term = T₁₉ = a + 18d …(iii)

Putting the value of equation (i) into equation (ii) and (iii)

we get

T₂₉ = (a + 8 d) + 20 d

⇒ T₂₉ = 0 + 20 d

⇒ T₂₉ = 20d … (iv)

⇒ T₂₉= (a + 8 d) + 10 d

⇒ T₂₉ = 0 + 10d

⇒ T₂₉ = 10d … (v)

From equation (iv) and (v), we have

T₂₉ = 20d = 2 × 10d =2 × T₂₉

T₂₉ = 2T₂₉

Hence Proved.